Deformations of K3 surfaces and orientation - users.unimi.itusers.unimi.it/stellari/Research/Slides/LucidiK3DefUtahBeamer.pdf · Deformations of K3 surfaces and orientation Paolo

Deformationsof K3

surfaces andorientation

Paolo StellariDeformations of K3 surfaces and

orientation

Paolo Stellari

Dipartimento di Matematica “F. Enriques”Universita degli Studi di Milano

Joint work with: E. Macrı (arXiv:0804.2552) and D. Huybrechts and E. Macrı (arXiv:0710.1645)

Deformationsof K3


Paolo Stellari

Outline

Outline

1 MotivationsThe settingThe problemThe analogies

2 Infinitesimal Derived Torelli TheoremThe settingThe statementSketch of the proof

3 OrientationThe statementThe strategyThe categorical settingDeforming kernelsConcluding the argument

Deformationsof K3


Paolo Stellari

Outline

Outline




Deformationsof K3


Paolo Stellari

Outline

Outline




Deformationsof K3


Paolo Stellari

MotivationsThe setting

The problem

The analogies

InfinitesimalDerivedTorelliTheoremThe setting

The statement

Sketch of the proof

OrientationThe statement

The strategy

The categoricalsetting

Deforming kernels

Concluding theargument

Outline




Deformationsof K3


Paolo Stellari


The problem

The analogies


The statement

Sketch of the proof


The strategy


Deforming kernels


Derived categories and K3 surfaces

Let X be a smooth projective complex variety. Denote byCoh(X ) the abelian category of coherent sheaves on X .

The main algebraic invariant we are going to study is thebounded derived category of coherent sheaves

Db(X ) := Db(Coh(X )).

A K3 surface is a smooth compact Kahler (complex)surface X such that:

X is simply connected.

The canonical bundle KX is trivial.

Deformationsof K3


Paolo Stellari


The problem

The analogies


The statement

Sketch of the proof


The strategy


Deforming kernels









Deformationsof K3


Paolo Stellari


The problem

The analogies


The statement

Sketch of the proof


The strategy


Deforming kernels









Deformationsof K3


Paolo Stellari


The problem

The analogies


The statement

Sketch of the proof


The strategy


Deforming kernels









Deformationsof K3


Paolo Stellari


The problem

The analogies


The statement

Sketch of the proof


The strategy


Deforming kernels









Deformationsof K3


Paolo Stellari


The problem

The analogies


The statement

Sketch of the proof


The strategy


Deforming kernels









Deformationsof K3


Paolo Stellari


The problem

The analogies


The statement

Sketch of the proof


The strategy


Deforming kernels


Outline




Deformationsof K3


Paolo Stellari


The problem

The analogies


The statement

Sketch of the proof


The strategy


Deforming kernels


The problem

Let X be a K3 surface.

Main problemDescribe the group of exact autoequivalences of thetriangulated category Db(X ) or of a first order deformation ofit.

Remark (Orlov)Such a description is available (in the non-deformedcontext) when X is an abelian surface (actually an abelianvariety).

Deformationsof K3


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The problem

The analogies


The statement

Sketch of the proof


The strategy


Deforming kernels


The problem




Deformationsof K3


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The problem

The analogies


The statement

Sketch of the proof


The strategy


Deforming kernels


The problem




Deformationsof K3


Paolo Stellari


The problem

The analogies


The statement

Sketch of the proof


The strategy


Deforming kernels


The problem




Deformationsof K3


Paolo Stellari


The problem

The analogies


The statement

Sketch of the proof


The strategy


Deforming kernels


Outline




Deformationsof K3


Paolo Stellari


The problem

The analogies


The statement

Sketch of the proof


The strategy


Deforming kernels


Geometry: automorphisms

Theorem (Torelli Theorem)Let X and Y be K3 surfaces. Suppose that there exists aHodge isometry

g : H2(X , Z) → H2(Y , Z)

which maps the class of an ample line bundle on X into theample cone of Y . Then there exists a unique isomorphismf : X ∼−→ Y such that f∗ = g.

Lattice theory + Hodge structures + ample cone

RemarkThe automorphism is uniquely determined.

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Theorem (Torelli Theorem)Let X and Y be K3 surfaces.

Suppose that there exists aHodge isometry

g : H2(X , Z) → H2(Y , Z)




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g : H2(X , Z) → H2(Y , Z)

which maps the class of an ample line bundle on X into theample cone of Y .

Then there exists a unique isomorphismf : X ∼−→ Y such that f∗ = g.



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g : H2(X , Z) → H2(Y , Z)




Deformationsof K3


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The problem

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The strategy


Deforming kernels




g : H2(X , Z) → H2(Y , Z)


Lattice theory

+ Hodge structures + ample cone


Deformationsof K3


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The analogies


The statement

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The strategy


Deforming kernels




g : H2(X , Z) → H2(Y , Z)


Lattice theory + Hodge structures

+ ample cone


Deformationsof K3


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The problem

The analogies


The statement

Sketch of the proof


The strategy


Deforming kernels




g : H2(X , Z) → H2(Y , Z)




Deformationsof K3


Paolo Stellari


The problem

The analogies


The statement

Sketch of the proof


The strategy


Deforming kernels




g : H2(X , Z) → H2(Y , Z)




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The analogies


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Deforming kernels


Geometry: diffeomorphisms

Theorem (Borcea, Donaldson)Consider the natural map

ρ : Diff(X ) −→ O(H2(X , Z)).

Then im (ρ) = O+(H2(X , Z)), where O+(H2(X , Z)) is thegroup of orientation preserving isometries.

The orientation is given by the choice of a basis for the3-dimensional positive space in H2(X , R).

RemarkThe kernel of ρ is not known!

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ρ : Diff(X ) −→ O(H2(X , Z)).




Deformationsof K3


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The problem

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The strategy


Deforming kernels




ρ : Diff(X ) −→ O(H2(X , Z)).




Deformationsof K3


Paolo Stellari


The problem

The analogies


The statement

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The strategy


Deforming kernels




ρ : Diff(X ) −→ O(H2(X , Z)).




Deformationsof K3


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Deforming kernels


Orlov’s result

Derived Torelli Theorem (Mukai, Orlov)Let X and Y be smooth projective K3 surfaces. Then thefollowing are equivalent:

1 There exists an equivalence Φ : Db(X ) ∼= Db(Y ).2 There exists a Hodge isometry H(X , Z) ∼= H(Y , Z).

The equivalence Φ induces an action on cohomology

Db(X )

v(−)=ch(−)·√

td(X)

Φ // Db(Y )

v(−)=ch(−)·√

td(Y )

H(X , Z)ΦH // H(Y , Z)

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Orlov’s result




Db(X )

v(−)=ch(−)·√

td(X)

Φ // Db(Y )

v(−)=ch(−)·√

td(Y )

H(X , Z)ΦH // H(Y , Z)

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Deforming kernels


Orlov’s result


1 There exists an equivalence Φ : Db(X ) ∼= Db(Y ).

2 There exists a Hodge isometry H(X , Z) ∼= H(Y , Z).


Db(X )

v(−)=ch(−)·√

td(X)

Φ // Db(Y )

v(−)=ch(−)·√

td(Y )

H(X , Z)ΦH // H(Y , Z)

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Orlov’s result




Db(X )

v(−)=ch(−)·√

td(X)

Φ // Db(Y )

v(−)=ch(−)·√

td(Y )

H(X , Z)ΦH // H(Y , Z)

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The problem

The analogies


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Deforming kernels


Orlov’s result




Db(X )

v(−)=ch(−)·√

td(X)

Φ // Db(Y )

v(−)=ch(−)·√

td(Y )

H(X , Z)ΦH // H(Y , Z)

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Main problem

QuestionCan we understand better the action induced oncohomology by an equivalence?

Orientation: Let σ be a generator of H2,0(X ) and ω aKahler class. Then 〈Re(σ), Im(σ), 1− ω2/2, ω〉 is a positivefour-space in H(X , R) with a natural orientation.

ProblemThe isometry j := (id)H0⊕H4 ⊕ (− id)H2 is not orientationpreserving. Is it induced by an autoequivalence?

Deformationsof K3


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The strategy


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Main problem




Deformationsof K3


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The problem

The analogies


The statement

Sketch of the proof


The strategy


Deforming kernels


Main problem




Deformationsof K3


Paolo Stellari


The problem

The analogies


The statement

Sketch of the proof


The strategy


Deforming kernels


Main problem




Deformationsof K3


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The problem

The analogies


The statement

Sketch of the proof


The strategy


Deforming kernels


Motivation

There exists an explicit description of the first orderdeformations of the abelian category of coherent sheaveson a smooth projective variety (Toda).

The existence of equivalences between the derivedcategories of smooth projective K3 surfaces is detected bythe existence of special isometries of the totalcohomologies.

QuestionCan we get the same result for derived categories of firstorder deformations of K3 surfaces using special isometriesbetween ‘deformations’ of the Hodge and lattice structureson the total cohomologies?

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The problem

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The strategy


Deforming kernels


Motivation




Deformationsof K3


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The problem

The analogies


The statement

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The strategy


Deforming kernels


Motivation




Deformationsof K3


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The problem

The analogies


The statement

Sketch of the proof


The strategy


Deforming kernels


Motivation




Deformationsof K3


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The problem

The analogies


The statement

Sketch of the proof


The strategy


Deforming kernels


Outline




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Hochschild homology and cohomology

For X any smooth projective variety, define the Hochschildhomology

HHi(X ) := Hom Db(X×X)(∆∗ω∨X [i − dim(X )],O∆X )

and the Hochschild cohomology

HHi(X ) := Hom Db(X×X)(O∆X ,O∆X [i]).

On the other hand we put

HΩi(X ) :=⊕

q−p=i

Hp(X ,ΩqX ) HTi(X ) :=

⊕p+q=i

Hp(X ,∧qTX ).

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HΩi(X ) :=⊕

q−p=i


⊕p+q=i

Hp(X ,∧qTX ).

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HΩi(X ) :=⊕

q−p=i


⊕p+q=i

Hp(X ,∧qTX ).

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Hochschild–Kostant–Rosenberg

There exist (the Hochschild–Kostant–Rosenberg)isomorphisms

IXHKR : HH∗(X ) → HΩ∗(X ) :=

⊕i

HΩi(X )

andIHKRX : HH∗(X ) → HT∗(X ) :=

⊕i

HTi(X ).

One then defines the graded isomorphisms

IXK = (td(X )1/2 ∧ (−)) IX

HKR IKX = (td(X )−1/2y(−)) IHKR

X .

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IXHKR : HH∗(X ) → HΩ∗(X ) :=

⊕i

HΩi(X )

and

IHKRX : HH∗(X ) → HT∗(X ) :=

⊕i

HTi(X ).


IXK = (td(X )1/2 ∧ (−)) IX


X .

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IXHKR : HH∗(X ) → HΩ∗(X ) :=

⊕i

HΩi(X )


⊕i

HTi(X ).


IXK = (td(X )1/2 ∧ (−)) IX


X .

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Deforming kernels




IXHKR : HH∗(X ) → HΩ∗(X ) :=

⊕i

HΩi(X )


⊕i

HTi(X ).


IXK = (td(X )1/2 ∧ (−)) IX


X .

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Toda’s construction

1 Take a smooth projective variety X , v ∈ HH2(X ) andwrite

IHKRX (v) = (α, β, γ) ∈ HT2(X ).

2 Define a sheaf O(β,γ)X of C[ε]/(ε2)-algebras on X

depending only on β and γ.

3 Representing α ∈ H2(X ,OX ) as a Cech 2-cocycleαijk one has an element α := 1− εαijk which is aCech 2-cocycle with values in the invertible elements ofthe center of O(β,γ)

X .

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IHKRX (v) = (α, β, γ) ∈ HT2(X ).




X .

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IHKRX (v) = (α, β, γ) ∈ HT2(X ).




X .

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IHKRX (v) = (α, β, γ) ∈ HT2(X ).




X .

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We get the abelian category

Coh(O(β,γ)X , α)

of α-twisted coherent O(β,γ)X -modules. Set

Coh(X , v) := Coh(O(β,γ)X , α).

One also have an isomorphism J : HH2(X1) → HH2(X1)such that

(IHKRX1

J (IHKRX1

)−1)(α, β, γ) = (α,−β, γ).

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Coh(O(β,γ)X , α)




(IHKRX1

J (IHKRX1

)−1)(α, β, γ) = (α,−β, γ).

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Coh(O(β,γ)X , α)




(IHKRX1

J (IHKRX1

)−1)(α, β, γ) = (α,−β, γ).

Deformationsof K3


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The problem

The analogies


The statement

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The strategy


Deforming kernels


Outline




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The strategy


Deforming kernels


The Infinitesimal Derived Torelli Theorem

Theorem (Macrı–S.)Let X1 and X2 be smooth complex projective K3 surfacesand let vi ∈ HH2(Xi), with i = 1, 2. Then the following areequivalent:

1 There exists a Fourier–Mukai equivalence

ΦeE : Db(X1, v1)∼−→ Db(X2, v2)

with E ∈ Dperf(X1 × X2,−J(v1) v2).

2 There exists an orientation preserving effective Hodgeisometry

g : H(X1, v1, Z)∼−→ H(X2, v2, Z).

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ΦeE : Db(X1, v1)∼−→ Db(X2, v2)



g : H(X1, v1, Z)∼−→ H(X2, v2, Z).

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ΦeE : Db(X1, v1)∼−→ Db(X2, v2)



g : H(X1, v1, Z)∼−→ H(X2, v2, Z).

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The structures

For X a K3, v ∈ HH2(X ) and σX is a generator for HH2(X ),let

w := IXK (σX ) + εIX

K (σX v) ∈ H(X , Z)⊗ C[epsilon]/(ε2).

The free Z[ε]/(ε2)-module of finite rank H(X , Z)⊗ Z[ε]/(ε2)is endowed with:

1 The Z[ε]/(ε2)-linear extension of the generalized Mukaipairing 〈−,−〉M .

2 A weight-2 decomposition on H(X , Z)⊗ C[ε]/(ε2)

H2,0(X , v) := C[ε]/(ε2) · w H0,2(X , v) := H2,0(X , v)

and H1,1(X , v) := (H2,0(X , v)⊕ H0,2(X , v))⊥.

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The structures







H2,0(X , v) := C[ε]/(ε2) · w H0,2(X , v) := H2,0(X , v)

and H1,1(X , v) := (H2,0(X , v)⊕ H0,2(X , v))⊥.

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The structures







H2,0(X , v) := C[ε]/(ε2) · w H0,2(X , v) := H2,0(X , v)

and H1,1(X , v) := (H2,0(X , v)⊕ H0,2(X , v))⊥.

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The structures







H2,0(X , v) := C[ε]/(ε2) · w H0,2(X , v) := H2,0(X , v)

and H1,1(X , v) := (H2,0(X , v)⊕ H0,2(X , v))⊥.

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The structures







H2,0(X , v) := C[ε]/(ε2) · w H0,2(X , v) := H2,0(X , v)

and H1,1(X , v) := (H2,0(X , v)⊕ H0,2(X , v))⊥.

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The structures

This gives the infinitesimal Mukai lattice of X with respect tov , which is denoted by H(X , v , Z).

An isometry

g : H(X1, v1, Z)∼−→ H(X2, v2, Z)

which is g = g0 ⊗ Z[ε]/(ε2), where g0 is an Hodge isometryof the Mukai lattices is called effective.

An effective isometry is orientation preserving if g0preserves the orientation of the four-space.

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An isometry

g : H(X1, v1, Z)∼−→ H(X2, v2, Z)



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The structures


An isometry

g : H(X1, v1, Z)∼−→ H(X2, v2, Z)



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An isometry

g : H(X1, v1, Z)∼−→ H(X2, v2, Z)



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Outline




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Deformations

We just sketch of the implication (i)⇒(ii).

Let ΦeE : Db(X1, v1)∼−→ Db(X2, v2) be an equivalence

with kernel E ∈ Dperf(X1 × X2,−J(v1) v2).

One shows that the restriction E ∈ Db(X1 × X2) of E isthe kernel of a Fourier–Mukai equivalenceΦE : Db(X1)

∼−→ Db(X2).

Using Orlov’s result, take the Hodge isometryg0 := (ΦE)H : H(X1, Z) → H(X2, Z).

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Deformations





∼−→ Db(X2).


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Deformations





∼−→ Db(X2).


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Deformations





∼−→ Db(X2).


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Deformations





∼−→ Db(X2).


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The isometry

Toda: since E is a first order deformation of E ,

(ΦE)HH(v1) = v2.

Important!Assume we know that any Hodge isometry induced by anequivalence Db(X1) ∼= Db(X2) is orientation preserving.

To conclude and prove that

g := g0 ⊗ Z[ε]/(ε2) : H(X1, v1, Z) → H(X2, v2, Z)

is an effective orientation preserving Hodge isometry, weneed two commutative diagrams.

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The isometry


(ΦE)HH(v1) = v2.



g := g0 ⊗ Z[ε]/(ε2) : H(X1, v1, Z) → H(X2, v2, Z)


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The isometry


(ΦE)HH(v1) = v2.



g := g0 ⊗ Z[ε]/(ε2) : H(X1, v1, Z) → H(X2, v2, Z)


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The isometry


(ΦE)HH(v1) = v2.



g := g0 ⊗ Z[ε]/(ε2) : H(X1, v1, Z) → H(X2, v2, Z)


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Commutativity I

Any Fourier–Mukai functor acts on Hochschild homology.

Theorem (Macrı–S.)Let X1 and X2 be smooth complex projective varieties andlet E ∈ Db(X1 × X2). Then the following diagram

HH∗(X1)(ΦE)HH //

IX1K

HH∗(X2)

IX2K

H(X1, C)(ΦE )H // H(X2, C)

commutes.

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Commutativity I



HH∗(X1)(ΦE)HH //

IX1K

HH∗(X2)

IX2K

H(X1, C)(ΦE )H // H(X2, C)

commutes.

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Commutativity I



HH∗(X1)(ΦE)HH //

IX1K

HH∗(X2)

IX2K

H(X1, C)(ΦE )H // H(X2, C)

commutes.

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The problem

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Deforming kernels


Commutativity II

Using that for K3 surfaces H0,2 is 1-dimensional and theprevious result, one get the following commutative diagram(for a Fourier–Mukai equivalence ΦE ):

HH∗(X1)(ΦE)HH

//

(−)σX1

HH∗(X2)

(−)(ΦE )HH(σX1)

HH∗(X1)

(ΦE)HH //

IX1K

HH∗(X2)

IX2K

H(X1, C)(ΦE)H // H(X2, C),

where σX1 is a generator of HH2(X1).

Deformationsof K3


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The problem

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The statement

Sketch of the proof


The strategy


Deforming kernels


Commutativity II


HH∗(X1)(ΦE)HH

//

(−)σX1

HH∗(X2)

(−)(ΦE )HH(σX1)

HH∗(X1)

(ΦE)HH //

IX1K

HH∗(X2)

IX2K

H(X1, C)(ΦE)H // H(X2, C),


Deformationsof K3


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The problem

The analogies


The statement

Sketch of the proof


The strategy


Deforming kernels


Commutativity II


HH∗(X1)(ΦE)HH

//

(−)σX1

HH∗(X2)

(−)(ΦE )HH(σX1)

HH∗(X1)

(ΦE)HH //

IX1K

HH∗(X2)

IX2K

H(X1, C)(ΦE)H // H(X2, C),


Deformationsof K3


Paolo Stellari


The problem

The analogies


The statement

Sketch of the proof


The strategy


Deforming kernels


Outline




Deformationsof K3


Paolo Stellari


The problem

The analogies


The statement

Sketch of the proof


The strategy


Deforming kernels


The motivation

We go back to the original problem of describing the groupof exact autoequivalences of the derived category of a K3surface.

Remarks1 To conclude the previous argument involving (first

order) deformations, we need to prove that anyequivalence induces an orientation preserving Hodgeisometry.

2 The (quite involved) proof of this result will usedeformation of kernels.

Deformationsof K3


Paolo Stellari


The problem

The analogies


The statement

Sketch of the proof


The strategy


Deforming kernels


The motivation





Deformationsof K3


Paolo Stellari


The problem

The analogies


The statement

Sketch of the proof


The strategy


Deforming kernels


The motivation


Remarks

1 To conclude the previous argument involving (firstorder) deformations, we need to prove that anyequivalence induces an orientation preserving Hodgeisometry.


Deformationsof K3


Paolo Stellari


The problem

The analogies


The statement

Sketch of the proof


The strategy


Deforming kernels


The motivation





Deformationsof K3


Paolo Stellari


The problem

The analogies


The statement

Sketch of the proof


The strategy


Deforming kernels


The motivation





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Paolo Stellari


The problem

The analogies


The statement

Sketch of the proof


The strategy


Deforming kernels


The statement

Main Theorem (Huybrechts–Macrı–S.)

Given a Hodge isometry g : H(X , Z) → H(Y , Z), then thereexists and equivalence Φ : Db(X ) → Db(Y ) such thatg = ΦH if and only if g is orientation preserving.

Szendroi’s Conjecture is true: In terms ofautoequivalences, this yields a surjective morphism

Aut (Db(X )) O+(H(X , Z)),

where O+(H(X , Z)) is the group of orientation preservingHodge isometries.

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Deforming kernels


The statement




Aut (Db(X )) O+(H(X , Z)),


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Deforming kernels


The statement




Aut (Db(X )) O+(H(X , Z)),


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The statement

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The strategy


Deforming kernels


The statement




Aut (Db(X )) O+(H(X , Z)),


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The statement

Sketch of the proof


The strategy


Deforming kernels


The ‘easy’ implication

The statement: If g is orientation preserving than it lifts toan equivance.

A result of Hosono–Lian–Oguiso–Yau (heavily relayingon Mukai/Orlov’s Derived Torelli Theorem) shows that,up to composing with the isometry j , every isometrycan be lifted to an equivalence.

Since we know that j is not orientation preserving weconclude using the following:

Remark (Huybrechts-S.)All known equivalences (and autoequivalences) areorientation preserving.

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The analogies


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The strategy


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Deformationsof K3


Paolo Stellari


The problem

The analogies


The statement

Sketch of the proof


The strategy


Deforming kernels







Deformationsof K3


Paolo Stellari


The problem

The analogies


The statement

Sketch of the proof


The strategy


Deforming kernels







Deformationsof K3


Paolo Stellari


The problem

The analogies


The statement

Sketch of the proof


The strategy


Deforming kernels







Deformationsof K3


Paolo Stellari


The problem

The analogies


The statement

Sketch of the proof


The strategy


Deforming kernels


Outline




Deformationsof K3


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The problem

The analogies


The statement

Sketch of the proof


The strategy


Deforming kernels


The non-orientation Hodge isometry

Take any projective K3 surface X .

Consider the non-orientation preserving Hodgeisometry

j := (id)H0⊕H4 ⊕ (− id)H2 .

Since one implication is already true, to prove the maintheorem, it is enough to show that j is not induced by aFourier–Mukai equivalence.

We proceed by contradiction assuming that there existsE ∈ Db(X × X ) such that (ΦE)H = j .

Deformationsof K3


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The analogies


The statement

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The strategy


Deforming kernels





j := (id)H0⊕H4 ⊕ (− id)H2 .



Deformationsof K3


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The problem

The analogies


The statement

Sketch of the proof


The strategy


Deforming kernels





j := (id)H0⊕H4 ⊕ (− id)H2 .



Deformationsof K3


Paolo Stellari


The problem

The analogies


The statement

Sketch of the proof


The strategy


Deforming kernels





j := (id)H0⊕H4 ⊕ (− id)H2 .



Deformationsof K3


Paolo Stellari


The problem

The analogies


The statement

Sketch of the proof


The strategy


Deforming kernels





j := (id)H0⊕H4 ⊕ (− id)H2 .



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The problem

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The statement

Sketch of the proof


The strategy


Deforming kernels


The idea of the proof

Huybrechts–Macrı–S.: For some particular K3surfaces we know that j is not induced by anyFourier–Mukai equivalence: K3 surfaces with trivialPicard group.

Deform the K3 surface in the moduli space such thatgenerically we recover the behaviour of a generic K3surface.

Deform the kernel of the equivalence accordingly.

Derive a contradiction using the generic case.

Deformationsof K3


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The problem

The analogies


The statement

Sketch of the proof


The strategy


Deforming kernels







Deformationsof K3


Paolo Stellari


The problem

The analogies


The statement

Sketch of the proof


The strategy


Deforming kernels







Deformationsof K3


Paolo Stellari


The problem

The analogies


The statement

Sketch of the proof


The strategy


Deforming kernels







Deformationsof K3


Paolo Stellari


The problem

The analogies


The statement

Sketch of the proof


The strategy


Deforming kernels







Deformationsof K3


Paolo Stellari


The problem

The analogies


The statement

Sketch of the proof


The strategy


Deforming kernels


Outline




Deformationsof K3


Paolo Stellari


The problem

The analogies


The statement

Sketch of the proof


The strategy


Deforming kernels


Formal deformations

Take R := C[[t ]] to be the ring of power series in t with fieldof fractions K := C((t)).

Define Rn := C[[t ]]/(tn+1). Then Spec (Rn) ⊂ Spec (Rn+1).

For X a smooth projective variety, a formal deformation is aproper formal R-scheme

π : X → Spf(R)

given by an inductive system of schemes Xn → Spec (Rn)(smooth and proper over Rn) and such that

Xn+1 ×Rn+1 Spec (Rn) ∼= Xn.

Deformationsof K3


Paolo Stellari


The problem

The analogies


The statement

Sketch of the proof


The strategy


Deforming kernels


Formal deformations




π : X → Spf(R)



Deformationsof K3


Paolo Stellari


The problem

The analogies


The statement

Sketch of the proof


The strategy


Deforming kernels


Formal deformations




π : X → Spf(R)



Deformationsof K3


Paolo Stellari


The problem

The analogies


The statement

Sketch of the proof


The strategy


Deforming kernels


Formal deformations




π : X → Spf(R)



Deformationsof K3


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The problem

The analogies


The statement

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The strategy


Deforming kernels


The categories

There exist sequences

Coh0(X ×R X ′) → Coh(X ×R X ′) → Coh((X ×R X ′)K )

Coh0(X ) → Coh(X ) → Coh((X )K )

where Coh0(X ×R X ′) and Coh0(X ) are the abeliancategories of sheaves supported on X × X and Xrespectively.

In this setting we also have the sequences

Db0(X ×R X ′) → Db

Coh(OX×RX ′-Mod) → Db((X ×R X ′)K )

Db0(X ) → Db

Coh(OX -Mod) → Db(XK )

Deformationsof K3


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The problem

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The statement

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The strategy


Deforming kernels


The categories






Db0(X ×R X ′) → Db


Db0(X ) → Db


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The categories






Db0(X ×R X ′) → Db


Db0(X ) → Db


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Deforming kernels


The key example: the twistor space

Let us focus now on the case when X is a K3 surface.

Definition

A Kahler class ω ∈ H1,1(X , R) is called very general if thereis no non-trivial integral class 0 6= α ∈ H1,1(X , Z) orthogonalto ω, i.e. ω⊥ ∩ H1,1(X , Z) = 0.

Take the twistor space X(ω) of X determined by the choiceof a very general Kahler class ω ∈ KX ∩ Pic (X )⊗ R:

π : X(ω) → P(ω).

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Definition



π : X(ω) → P(ω).

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Deforming kernels




Definition



π : X(ω) → P(ω).

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Deforming kernels




Definition



π : X(ω) → P(ω).

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Deforming kernels



RemarkX(ω) parametrizes the complex structures ‘compatible’ withω.

Choosing a local parameter t around 0 ∈ P(ω) we get aformal deformation X → Spf(R).

More precisely:

Xn := X(ω)× Spec (Rn),

form an inductive system and give rise to a formalR-scheme

π : X → Spf(R),

which is the formal neighbourhood of X in X(ω).

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Deforming kernels





More precisely:



π : X → Spf(R),


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The analogies


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The strategy


Deforming kernels





More precisely:



π : X → Spf(R),


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The analogies


The statement

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The strategy


Deforming kernels





More precisely:



π : X → Spf(R),


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The statement

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The strategy


Deforming kernels


Equivalences

As before, given F ∈ DbCoh(OX×RX ′-Mod), we denote by FK

the natural image in the category Db((X ×R X ′)K ).

Proposition

Let E ∈ Db(X ×R X ′) be such that E = i∗E . Then E and EKare kernels of Fourier–Mukai equivalences.

Here we denoted by i : X × X → X ×R X ′ the naturalinclusion.

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Equivalences



Proposition



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The problem

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Deforming kernels


Equivalences



Proposition



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Deforming kernels


Equivalences



Proposition



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Outline




Deformationsof K3


Paolo Stellari


The problem

The analogies


The statement

Sketch of the proof


The strategy


Deforming kernels


The first order deformation

The equivalence ΦE induces a morphim

ΦHHE : HH2(X ) → HH2(X ).

Proposition

Let v1 ∈ H1(X , TX ) be the Kodaira–Spencer class of firstorder deformation given by a twistor space X(ω) as above.Then

v ′1 := ΦHHE (v1) ∈ H1(X , TX ).

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Proposition


v ′1 := ΦHHE (v1) ∈ H1(X , TX ).

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Proposition


v ′1 := ΦHHE (v1) ∈ H1(X , TX ).

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Let X ′1 be the first order deformation corresponding to v ′1.

Using results of Toda one gets the following conclusion

Proposition (Toda)

For v1 and v ′1 as before, there exists E1 ∈ Db(X1 ×R1 X ′1)

such thati∗1E1 = E0 := E .

Here i1 : X0 ×C X0 → X ′1 ×R1 X ′

1 is the natural inclusion.

Hence there is a first order deformation of E .

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Proposition (Toda)



Here i1 : X0 ×C X0 → X ′1 ×R1 X ′



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Proposition (Toda)



Here i1 : X0 ×C X0 → X ′1 ×R1 X ′



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Proposition (Toda)



Here i1 : X0 ×C X0 → X ′1 ×R1 X ′



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Proposition (Toda)



Here i1 : X0 ×C X0 → X ′1 ×R1 X ′



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Higher order deformations

More generally

We construct, at any order n, a deformation X ′n such that

there exists En ∈ Db(Xn ×Rn X ′n), with

i∗nEn = En−1.

Main difficulties1 Write the obstruction to deforming complexes in terms

of Atiyah–Kodaira classes (Huybrechts–Thomas).2 Show that the obstruction is zero.

Our approach imitates the first order case (using relativeHochschild homology).

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More generally



i∗nEn = En−1.




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Deforming kernels



More generally



i∗nEn = En−1.

Main difficulties

1 Write the obstruction to deforming complexes in termsof Atiyah–Kodaira classes (Huybrechts–Thomas).

2 Show that the obstruction is zero.


Deformationsof K3


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The problem

The analogies


The statement

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The strategy


Deforming kernels



More generally



i∗nEn = En−1.


of Atiyah–Kodaira classes (Huybrechts–Thomas).

2 Show that the obstruction is zero.


Deformationsof K3


Paolo Stellari


The problem

The analogies


The statement

Sketch of the proof


The strategy


Deforming kernels



More generally



i∗nEn = En−1.




Deformationsof K3


Paolo Stellari


The problem

The analogies


The statement

Sketch of the proof


The strategy


Deforming kernels



More generally



i∗nEn = En−1.




Deformationsof K3


Paolo Stellari


The problem

The analogies


The statement

Sketch of the proof


The strategy


Deforming kernels


Outline




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Deforming kernels


The generic fiber

Use the generic analytic caseThere exist integers n and m such that the Fourier–Mukaiequivalence

Φn(I∆X [1])K

ΦEK [m]

has kernel GK ∈ Coh((X ×R X ′)K ), for G ∈ Coh(X ×R X ′).

RemarkThis shows that the autoequivalences of the derivedcategory Db(XK ) behaves like the derived category of acomplex K3 surface with trivial Picard group(Huybrechts–Macrı–S.).

Deformationsof K3


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The generic fiber


Φn(I∆X [1])K

ΦEK [m]



Deformationsof K3


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The generic fiber


Φn(I∆X [1])K

ΦEK [m]



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The conclusion

Properties of G1 G0 := i∗G is a sheaf in Coh(X × X ).2 The natural morphism

(ΦG0)H : H∗(X , Q) → H∗(X , Q)

is such that (ΦG0)H = (ΦE)H = j .

LemmaIf G0 ∈ Coh(X × X ), then (ΦG0)H 6= j .

Deformationsof K3


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The conclusion

Properties of G

1 G0 := i∗G is a sheaf in Coh(X × X ).2 The natural morphism

(ΦG0)H : H∗(X , Q) → H∗(X , Q)



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The conclusion

Properties of G1 G0 := i∗G is a sheaf in Coh(X × X ).

2 The natural morphism

(ΦG0)H : H∗(X , Q) → H∗(X , Q)



Deformationsof K3


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The problem

The analogies


The statement

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The strategy


Deforming kernels


The conclusion


(ΦG0)H : H∗(X , Q) → H∗(X , Q)



Deformationsof K3


Paolo Stellari


The problem

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The statement

Sketch of the proof


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Deforming kernels


The conclusion


(ΦG0)H : H∗(X , Q) → H∗(X , Q)



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Deformations of K3 surfaces and orientation - users.unimi.itusers.unimi.it/stellari/Research/Slides/LucidiK3DefUtahBeamer.pdf · Deformations of K3 surfaces and orientation Paolo